Every isosceles trapezoid has an inscribed circle I think it's not true but I'm not sure about it . Can we use Pitot theorem here ? For example a trapezoid with long bases and short legs can't have an inscribed circle . Please explain about the conditions for an isosceles trapezoid to have an inscribed circle .
 A: Consider this figure.

I claim that if one of those trapezoids has an inscribed circle, the other does not.
A: If a quadrilateral has an inscribed circle it has to satisfy $\overline{AB} + \overline{CD} = \overline{BC} + \overline{AD}$. This is Pitot's theorem. The converse is also true.
Let $ABCD$ be a quadrilateral satisfying $\overline{AB} + \overline{CD} = \overline{BC} + \overline{AD}$. If $ABCD$ is a parallelogram then it is a rhombus, which has an inscribed circle. Now suppose that $\overline{BC}$ is not parallel to $\overline{AD}$. Let $P$ be the intersection of the line $\overleftrightarrow{BC}$ and $\overleftrightarrow{AD}$. Then the triangle $\triangle PCD$ has a inscribed circle.
Find $A'$ on the segment $\overline{PC}$ and $B'$ on the segment $\overline{PD}$ so that $\overline{A'B'}$ is parallel to $\overline{AB}$. If you vary $A'$ and $B'$ under this condition, there is a unique pair of $\{A'_0, B'_0\}$ satisfying $\overline{A'_0B'_0} + \overline{CD} = \overline{A'_0D} + \overline{B'_0C}$, and there is a unique pair of $\{A'_1, B'_1\}$ satisfying $\overline{A'_1B'_1}$ is tangent to the inscribed circle. $\{A'_0, B'_0\}$ must coincide $\{A'_1, B'_1\}$.
A: If the parallel sides of the trapezium have lengths $a$ and $b$ then by consideration of the tangents to the inscribed circle, the non-parallel slant edges would have to have their lengths both equal to $$\frac{a+b}{2}$$
